Problem: Write the equation for a parabola with a focus at $(-2,5)$ and a directrix at $x=3$. $x=$
The strategy A parabola is defined as the set of all points that are the same distance away from a point (the focus) and a line (the directrix). Let $(x,y)$ be a point on the parabola. Then the distance between $(x,y)$ and the focus, $(-2,5)$, is equal to the distance between $(x,y)$ and the directrix, $x=3$. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${\llap{-}10}$ ${\llap{-}12}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $y$ $x$ $(x,y)$ $(-2,5)$ $x=3$ Once we find these distances, we can equate them in order to derive the equation of our parabola. Finding the distances from $(x,y)$ to the focus and the directrix The distance between $(x,y)$ and $(-2,5)$ is $\sqrt{(x+2)^2+(y-5)^2}$. [How did we find that?] Similarly, the distance between $(x,y)$ and the line $x=3$ is $\sqrt{(x-3)^2}$. [How did we know that?] Deriving the formula by equating the distances $\begin{aligned} \sqrt{(x-3)^2} &= \sqrt{(x+2)^2+(y-5)^2} \\\\ (x-3)^2 &= (x+2)^2+(y-5)^2 \\\\ {x^2}-6x{+9} &= {x^2}{+4x}+4+(y-5)^2\\\\ -6x{-4x}&=(y-5)^2+4{-9} \\\\ -10x&=(y-5)^2-5 \\\\ x&=-\dfrac{(y-5)^2}{10}+\dfrac{1}{2}\end{aligned}$ The answer The equation of our parabola is $x=-\dfrac{(y-5)^2}{10}+\dfrac{1}{2}$. [Is the equation in vertex form?]